Answer:
Specific heat transfer = 236.16 kJ/kg
Ratio of return velocity to inlet velocity = 0.80
Explanation:
Given
Temperature of liquid nitrogen, T1 = 90 K
Pressure of liquid nitrogen, P1 = 400 kPa
Temperature of nitrogen, T2 = 160 K
Pressure of nitrogen, T2 = 400 kPa
A(e) = 100 A(i)
To solve, we use the formula
h(i)+ 1/2v(i)² + q = h(e) + 1/2v(e)² + q
The mass flow is
m = m(i) = m(e)
m = (Av/V)i = (Av/V)e
Ratio of return velocity to inlet velocity is
v(e) / v(i) = A(i)/A(e) * V(e)/V(i)
v(e) / v(i) = 1/100 * V(e)/V(i)
From the saturated Nitrogen table, at 100 K, we have
h(i) = h(f) = -73.2
v(i) = v(f) = 0.001452
From the saturated Nitrogen table again, at 160 K and 400 kPa
h(e) = 162.96 kJ/kg
v(e) = 0.11647 m³/kg
Substituting these in the formula, we have
v(e) / v(i) = 1/100 * 0.11647/0.001452
v(e) / v(i) = 1/100 * 80.2
v(e) / v(i) = 0.80
Energy equation is given by
q + h(i) = h(e)
q = h(e) - h(i)
Now, calculating specific heat transfer
q = 162.96 - -73.2
q = 236.16 kJ/kg
